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Problem 12.1: The set point of the control system shown in the figure below is given a

step change of 0.1unit in set point.

a) Develop an overall transfer function)(

)(

sR

sC

b) Determine the order of the overall system. If it is second order system, determine thevalues of andand state whether the system is underdamped, critically damped oroverdamped.

c) Calculate the ultimate value of the response )(tC

d) Calculate the offset

e) Calculate the maximum value of )(tC and the time at which it occurs

f) Find out the period of oscillation

Draw sketch ofC(t) as a function of time.

Solution:

)12)(1(

56.11

)12)(1(

56.1

)(

)(

++

+

++

=

ss

ss

sR

sC

932

8

132

8132132

8

)12)(1(

81

)12)(1(

8

)(

)(2

2

2

2

++=

+++++

++=

+++

++=ss

ss

ss

ss

ss

ss

sR

sC

932

8

)(

)(2

++

=

sssR

sC[solution to the part (a)] (1)

For second order system, we know

12)(

)(22 ++

=ss

K

sX

sY

(2)

Eq. (1) can also be written as

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13

1

9

2

9/8

9

9

9

3

9

2

9/8

9

932

9/8

932

8

)(

)(

2222

++=

++=

++=

++=

sssssssssR

sC

(3)

Comparing Eq. 1 and Eq. 3, it may be shown that Eq. 1 is a transfer function of a secondorder system [solution to part (b)], and we may write

9

8=K

9

22 =

3

2= [solution to part (b)] (4)

3

12 =

354.022

1

2

3

2

1

3

1=== [solution to part (b)] (5)

As 0.1

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0111.00889.01.0 ==Offset [solution to part (d)]

+

=

tteAKtY

t

2

2

2 1sin

1

1cos1)(

Time to first peak: 21

=pt

Overshoot:

==

21exp

B

AOS

Decay ratio:

===

2

2

1

2exp)(

OS

A

CDR

Time period: 21

2

=T

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Solution:

a)

+

+

++

+

+=

1

1

1

1)1(1

1

1)1(

)(

)(

1

1

sssK

ssK

sR

sC

m

Dc

Dc

)1()1()1(

)1()1(

)1).(1(

)1()1()1(

1

1)1(

1

1

1

1

sKss

ssK

ss

sKss

ssK

Dcm

mDc

m

Dcm

Dc

++++

++

=++

++++

+

+

=

sKKsss

ssK

Dccmm

mDc

+++++++

=1

)1()1(

1

2

1

)1()(

)1()1(

1

2

1 +++++++

=cDcmm

mDc

KsKs

ssK

1)1(

)(

)1(

)1(1)1()1(

)(

)(

121 ++

+++

+

+++=

sK

Ks

K

KssK

sR

sC

c

Dcm

c

m

c

mDc

This is resembling to a second order system, therefore

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11

112

+=

+=

c

m

c

m

KK

)1(

)(2 1

+++

=c

Dcm

K

K

)1(

)(1

2

1 1

+++

=c

Dcm

K

K

)1(

)(

1

1

2

1 1

1+

++

+

=c

Dcm

c

mK

K

K

)()1(

1

2

11

1

Dcm

cm

KK

+++

=

b) When 0=D s

)60

101(

)1(60

101

1

2

17.0 +

+=

cK

)1(6

1

60/35

60

1060

)1(6

1

1

2

17.0

+

=

+

+

=

cc KK

167.3=cK

b) When 3=D s

++

+=

60

3

60

101

)1(6

1

1

2

17.0 c

c

K

K

+

+=

60

370

)1(6

1

1

2

17.0 c

c

K

K

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07.060

370

)1(6

1

1

2

1 =

+

+ c

c

K

K

Here, we can use Solver (in MicroSoft Excel).

255.5=cK

Write the following formula in D7

=(0.5*(1/SQRT((1/6)*(D5+1)))*((70+3*D5)/60))-0.7

Go to Tools then Solver. Click Solver.

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In the Solver window below, do the following:

Set Target Cell: D7; Select Equal To: value of: 0; By Changing Cells: D5

ClickSolve button.

The solution is there.

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c)

+

+

++

+

+=

1

1

1

1)1(1

1

1)1(

)(

)(

1

1

sssK

ssK

sR

sC

m

Dc

Dc

For a unit step change:

ssR

1)( =

+

+

++

+

+=

1

1

1

1)1(1

1

1)1(

1)(

1

1

sssK

ssK

ssC

m

Dc

Dc

For the ultimate value, t , and )()( CtC . Now using final value theorem,

)]([0

lim)]([

limssf

stf

t =

c

c

K

KC

+=1

)(

When 167.3=cK

760.0167.31

167.3)( =

+

=C and Offset 24.0760.01 ==

When 255.5=cK

840.0255.51

255.5)( =

+

=C and Offset 160.0840.01 ==

Please compare the above two offsets and also find periods and compare.

Period is greater when 0=D .

Stability of a Control Loop

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Consider the following block diagram

Taking 11 = , ,2

12 = and

3

13 = , find the response C(t) for a step change in the set

point i.e.s

sR1

)( = .

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Definition of stability

A control system is said to be stable when the output response is bounded for all bounded

inputs.

Or

A process is said to be unstable if its output becomes larger and larger (either positivelyor negatively) as time increases.

A bounded inputis a function of time that always falls within certain bounds during thecourse of time.

Step function is a bounded input while the function ttf =)( is unbounded.

Saturation

If we keep on opening the valve, a point will reach where further pressure to the

diaphragm will not change the flowrate through the valve, such type of limitation iscalled saturation. Therefore physical systems saturate and do not go beyond bounds.

However, working at saturation is of course not a satisfactory control. With some

exothermic reactions, the saturation point may be highly unsatisfactory and may damagethe equipment and in some cases the human life. Also a saturated or near saturated value

of a variable may affect the quality of a product.

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Figure 1 Stable and unstable responses

Characteristic equation

Consider the following block diagram

Where,

cKsG =)(1

)1)(1(

1)(

21

2++

=

sssG

1

1

)( 3 +=

ssH

Overall transfer function for set point change

)()()(1

)()(

)(

)(

21

21

sHsGsG

sGsG

sR

sC

+

=

Overall transfer function for load point change

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)()()(1

)(

)(

)(

21

2

sHsGsG

sG

sU

sC

+=

The denominator is same in both cases, when equal to zero is called the characteristic

equation.

The term )()()( 21 sHsGsG in the denominator is called the open loop transfer

function.

Roots of the characteristic equation

For unit step change in set point, ssR

1

)( = , it may shown that

( ))1()()()1(

)(321

2323121

3321

3

c

c

Kssss

sKsC

++++++++

+=

))()((

)/()1()(

321

3213

rsrsrss

sKsC c

+

=

Where r1, r2, and r3 are the roots of the following characteristic equation.

0)1()()( 3212

3231213

321 =++++++++ cKsss

Criteria of stability and roots of a characteristic equation

The feedback control system is stable if and only if all roots of the characteristic equation

are negative or have negative real parts.

A linear control system is unstable if any roots of its characteristic equation are

On, or to the right of the imaginary axis. Otherwise system is stable.

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Figure: Stability criteria and roots of the characteristic equation

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Example 13.1 For the figure below, a control system has the transfer functions

Characteristic equation:

0)12(

)15.0(101 =+

++ ss

s

0532 =++ ss

One can solve the quadratic equation by hand as well. However, solving by Matlab

Since real parts ofs1 ands2 are negative, the system is stable.

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Routh test for stability

It is an algebraic method for determining how many roots of the characteristicequation have positive real pars (No roots with positive real parts, the system is stable).

The test is limited to the systems that have polynomial characteristic equations.Therefore it can not be used to test a control system with a transportation lag.

It gives no information about the actual location of the roots and, in particular,their proximity to the imaginary axis.

Procedure for examining the roots

o Write the characteristic equation in the form

02

2

1

10 =++++

n

nnn

asasasa

Where a0 is positive (ifa0 is originally negative, both sides are multiplied by 1).

o The coefficients of above equation should be positive. If any coefficient is negative,

the system is definitely unstable and Routh test is not needed.

o If all the coefficients are positive, the system may or may not be stable.

o Arrange the coefficients into the first two rows of the Routh array as shown in the

next slide.

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o For a stable system all elements of the first column of the Routh array should be

positive and nonzero.

o If some of the elements in the first column are negative, the number of roots with

positive real part (in the right half plane) is equal to the number of sign changes in the

first column.o If one pair of roots is on the imaginary axis, equidistant from the origin, and all other

roots are in the left half-plane, then all the elements of the nth row will vanish and

none of the elements of the preceding row will vanish. The location of the pair of

imaginary roots can be found by solving the equation

02 =+DCs

where the coefficients Cand D are the elements of the array in the (n1)st row as

read from left to right, respectively.

Example 13.2

Given the characteristic equation

02453234 =++++ ssss , determine the stability by the Routh criterion.

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Problem 13.1

Write the characteristic equation and construct the Routh array for the control systemshown in the figure below. Is the system stable for (a) Kc = 9.5, (b)Kc = 11, and (c)Kc =

12?

Overall transfer function:

+

+++

++

=

+

++

+

++

=

3

3

)12

)(1(

11

)12

)(1(

1

3

3

)15.0)(1(

11

)15.0)(1(

1

)(

)(

sss

K

ss

K

sssK

ssK

sR

sC

c

c

c

c

)3(

6)3)(2)(1(

2

)3)(2)(1(

61

)2)(1(

2

3

3

)2)(1(

2

1

)2)(1(

2

+

++++=

+++

+

++

=

+

+++

++

=

s

Ksss

K

sss

K

ss

K

sssK

ssK

c

c

c

c

c

c

c

c

Ksss

sK

sR

sC

++++

+=

6)3)(2)(1(

)3(2

)(

)(

Characteristic equation:

03

3

)15.0)(1(

11 =

+

++

+sss

Kc

03

3

)12

)(1(

11 =

+

+++

sss

Kc

03

3

)2)(1(

21 =

+

++

+sss

Kc

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0)3)(2)(1(

61 =

+++

+sss

Kc

06)3)(2)(1( =++++ cKsss

06)3)(23(2 =++++ cKsss

06611623 =++++ cKsss

0)1(6116 23 =++++ cKsss

Routh array:

Row1 1 11

2 6 )1(6 cK+

3 cK10

Note the calculations of forb1: ccc KKK

b =

=+

= 101

111

6

)1(611161

For 5.9=cK

5.05.91010 == cK

For 11=cK

1111010 ==c

K

For 12=cK

2121010 ==c

K

System is stable only forKc = 2.0.

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For 0.2=cK ,

01042

102

3121

2=

++

++

sss

0120504223

=+++ sss

Row

1 2 50

2 4 120

3 10System is unstable.

19